How do I find the sum of this?

[grizzled bane]

When I mean sum I mean x^n-1

[grizzled bane]

This isn't a homework question either btw

[left saddle]

$\sum_{n=1}^{\infty} x^{n-1} = \frac{1}{x} \sum_{n=1}^{\infty} x^{n}$

[tame hornet]

Yojda

[grizzled bane]

I think i remeber checking on wolfram alpha it was -ln(1-x)/x but i actually wanna learn how to

[left saddle]

Then find value of geometric sum

[grizzled bane]

Ok I get the factoring

How would I find sum of x^n? I know it has to be a power series

[left saddle]

Yup, but the formula is

[grizzled bane]

Also the step i showed in the beginning was a technique i saw from michael penn on yt

[left saddle]

$\sum_{n=1}^{\infty} x^{n} = \frac{1}{1-x} - 1$

[grizzled bane]

Give me a second..

[left saddle]

np

(for |x| < 1)

[grizzled bane]

Why does wolfram say this

[tame hornet]

Yojda

[left saddle]

Yes I forgot the -1

I was searching on google

Because it's the formula for n goes 0 to infty

[grizzled bane]

Ok I understand why its 1/1-x but why the extra 1

[left saddle]

$\sum_{k=0}^{p} x^{k} = \frac{1 - x^{p+1}}{1-x}$

[tame hornet]

Yojda

[left saddle]

So $\sum_{k=1}^{p} x^{k} = \frac{1 - x^{p+1}}{1-x} - x^0$

[tame hornet]

Yojda

[grizzled bane]

This is the formula when a geometric series diverges or something i think

[left saddle]

Okay. Series = Sum goes to infinity

And my formula is for finite sum

But when the sum goes to infinity, we have to look at x

[grizzled bane]

Not sure if i did it right

[grizzled bane]

So does this involve a limit

[left saddle]

On wolfram alpha is $\frac{-x}{x-1}$, not $\frac{-x}{1-x}$

[tame hornet]

Yojda

[left saddle]

for |x| < 1, it converges

[grizzled bane]

So its same as this?

[left saddle]

Yes

[grizzled bane]

Oh you just factor a negative right

[left saddle]

$\frac{1}{1-x} - 1 = \frac{1 - 1 + x}{1-x} = \frac{x}{1-x} = \frac{-x}{x-1}$

[tame hornet]

Yojda

[left saddle]

Oops, yes it's good

[grizzled bane]

Ok so far i still dont understand where 1 comes from

[left saddle]

$x^0$

[tame hornet]

Yojda

[left saddle]

I removed the case when n=0 from the sum

[grizzled bane]

I didn't ask before but the picture at thr beginning is that correct or did i do something wrong

Oh im tripping i meant x^n-1 /n

[left saddle]

This ?

[grizzled bane]

This i what im trying to achieve

My bad

This is what i meant at the beginning

Because integral from (0,1) of x^n-1 /n will give me 1/n^2

[grizzled bane]

Is the new sum complicated

Guess im gonna have to make a different thread

Thanks for the help

.close